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\title{《基础复分析》第7章调和函数 - 习题}
\author{CGZ ET AL}

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%## 《基础复分析》习题七

\begin{enumerate}

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\item % 1

设 $f(z)$ 在圆环 $r_1 < |z| < r_2$ 内解析，且在闭圆环上连续。
记 $M(r)$ 为 $|f(z)|$ 在圆周 $|z|=r$ 上的最大值。

证明：
$$
M(r) \leq M(r_1)^{\alpha} M(r_2)^{1-\alpha}
$$

其中

$$
\alpha = \frac{\ln r_2 - \ln r}{\ln r_2 - \ln r_1}
$$

并讨论等号成立的情形。
    
(提示：对 $\ln |f(z)|$ 和 $\ln |z|$ 的一个线性组合应用极值原理。)
    

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\item % 2

通过对 $\ln |1+z|$ 应用均值公式，计算积分
    $$
    \int_0^{\pi} \ln(\sin \theta) \, d\theta.
    $$
    

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\item % 3

设 $f(z)$ 在全平面解析，且 
$$
\lim_{z \to \infty} (\operatorname{Re} f(z))/z = 0$$

试证 $f(z)$ 是常数。

(提示: 利用 Schwarz 公式。)
    

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\item % 4

设 $f(z)$ 在 $\infty$ 的一个邻域内解析, 且 $\lim_{z \to \infty} (\operatorname{Re} f(z))/z = 0$. 

试证 $\lim_{z \to \infty} f(z)$ 存在.

(提示: 利用 Cauchy 积分公式证明 $f(z) = f_1(z) + f_2(z)$, 其中 $\lim_{z \to \infty} f_2(z) = 0$, 而 $f_1(z)$ 在全平面解析.)
    

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\item % 5

设 $u(z)$ 在 $0 < |z| < \rho$ 内调和且有界, 证明原点是 $u(z)$ 的可去奇点.
    

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\item % 6

设 $u(z)$ 在 $0 < |z| < \rho$ 内调和, 且 $\lim_{z \to 0} zu(z) = 0$. 试证存在常数 $\alpha$ 使得 $u(z) - \alpha \ln |z|$ 以原点为可去奇点.
    

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\item % 7

证明在上半平面调和且有界, 在实轴上连续的函数 $u(z)$ 可以表示为
    $$
    u(z) = \frac{1}{\pi} \int_{-\infty}^{+\infty} \frac{y}{(x-\xi)^2 + y^2} u(\xi) \, d\xi.
    $$
    

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\item % 8

设 $u(z)$ 在上半平面调和且有界, 在实轴上除去原点之外连续, 且 $\lim_{x \to 0-} u(x) = 1$, $\lim_{x \to 0+} u(x) = 0$. 证明
    $$
    \lim_{z \to 0} \left( u(z) - \frac{\arg z}{\pi} \right) = 0.
    $$
    

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\item % 9

设 $C_1, C_2$ 为单位圆周去掉两点后余下的两段开弧. 试求单位圆盘内的有界调和函数 $u(z)$, 使得
    $$
    \lim_{z \to \xi \in C_1} u(z) = 1, \quad \lim_{z \to \xi \in C_2} u(z) = 0.
    $$
    并证明与 $C_1$ 相对, 由过 $z$ 点以及 $C_1$ 的两端点的直线所截出的弧长度为 $2\pi u(z)$.
    

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\item % 10

设 $f(z)$ 在 $\mathbb{C}$ 上解析, 在实轴上取实值, 在虚轴上取纯虚数. 证明 $f(z)$ 是奇函数.
    

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\item % 11

证明: 关于实轴对称的区域 $\Omega$ 上的解析函数 $f(z)$, 可以分解为 $f(z) = f_1(z) + i f_2(z)$, 使得 $f_1, f_2$ 都在 $\Omega$ 上解析, 且在实轴上取实值.
    

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\item % 12

如果 $f(z)$ 在包含闭单位圆盘的区域内解析, 且把单位圆周映到自身, 证明 $f(z)$ 是有理函数.
    

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\item % 13

设 $u(z)$ 在上半平面调和, 且 $0 \leq u(z) \leq Ky$. 证明 $u(z) = ky$, 其中 $0 \leq k \leq K$.

(提示: 通过反射原理得到以 $u(z)$ 为实部的解析函数 $f(z)$, 再利用 Schwarz 公式证明 $f'(z)$ 是有界的.)

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\end{enumerate}

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